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I am trying to learn about free groups(as part of my Bachelor's thesis), and was assigned with Hungerford's Algebra book. Unfortunately, the book uses some aspects from category theory(which I have not learned). If someone has an access to the book and can help me, I would be grateful.

First, in theorem 9.1(2003 edition), he proves that the set of all reduced words, $F(X)$, of a given set $X$ is a group(under the operation defined), using "Van Der Waerden trick". At some point in the proof he says:" since $1\mapsto x_{1}^{\delta_{1}}x_{2}^{\delta_{2}}\cdots x_{n}^{\delta_{n}}$, under the map $|x_{1}^{\delta_{1}}|\cdots |x_{n}^{\delta_{n}}|$, then it follows that $\varphi$ is injective". Can you please explain how did he get this conclusion?

Second, he found that this $\varphi$ is a bijection between a set and a group, and he concludes that since the group is associative then also the set is. Why is that?

Thank you.

Mr. Tea
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  • I sympathize. I just pulled the book off my shelf and read the section in question. It is dense and much too formal for one's first exposure to this. I wish I had a better reference for you. – Randall Oct 19 '17 at 13:30
  • @Randall, Thank you for sympathy. I am just afraid to let my supervisor down by saying that it a bit difficult for me to get it all done in a week. – Mr. Tea Oct 19 '17 at 13:34
  • Do you have any good intuition about what $F(X)$ should be as a group, in a down-to-earth sense? If not, this is why you can't read it: his exposition is not going to help you learn what it is, only to prove (really, really rigorously) that it does what you want it to do as a free object. In some sense, free groups are the simplest groups conceptually, but getting the details right is hard (and not always worth it up front). – Randall Oct 19 '17 at 13:37
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    IMHO, this link is a better starting point than Hungerford: https://math.stackexchange.com/questions/1667282/precise-definition-of-free-group?rq=1 – Randall Oct 19 '17 at 13:39
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    I think that I have an idea about what $F(X)$ should be as a group, and I fully understood the construction of the set $F(X)$, and how the operation on the set works. In the theorem I have problem with, he is just trying to avoid the ugly proof(by induction) of the associative law on the group by finding a group in which induces its associative law on $F(x)$(hope I am not wrong). – Mr. Tea Oct 19 '17 at 13:49
  • Yes, he is (painfully) trying to prove that associativity holds in $F(X)$. Note than even he declares his proof as a "sketch." – Randall Oct 19 '17 at 13:55

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